In a survey of adults aged 57 through 85 years, it was found that 86.6% of them used at least one prescription medication. Complete parts (a) through (c) below.

a. How many of the 3149 subjects used at least one prescription medication?

(Round to the nearest integer as needed.)

b. Construct a 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication.

(Round to one decimal place as needed.)

b) The 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication is (85.6%, 87.6%).

Step-by-step explanation:

Question a:

86.6% out of 3149, so:

0.866*3149 = 2727.

272 used at least one prescription medication.

Question b:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 3149, \pi = 0.866[/tex]

90% confidence level

So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.866 - 1.645\sqrt{\frac{0.866*0.134}{3149}} = 0.856[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.866 + 1.645\sqrt{\frac{0.866*0.134}{3149}} = 0.876[/tex]

For the percentage:

0.856*100% = 85.6%

0.876*100% = 87.6%.

The 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication is (85.6%, 87.6%).

astha8579 astha8579 · Answer 2 · 2022-03-30T08:05:55+02:00

The number of subjects in the study who used at least one prescription medication is 2727 approx. The needed 90% confidence interval is: [0.8560, 0.8758] or in percentage as: 85.6% to 87.58%

How to construct confidence interval for population proportion based on the sample proportion?

Suppose that we have:

n = sample size
[tex]\hat{p}[/tex] = sample proportion
[tex]\alpha[/tex] = level of significance = 1 - confidence interval = 100 - confidence interval in percentage

Then, we get:

[tex]CI = \hat{p} \pm Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

where [tex]Z_{\alpha/2}[/tex] is the critical value of Z at specified level of significance and is obtainable from its critical value table(available online or in some books)

For this case, we have:

n = 3149
confidence interval is of 90%
[tex]\alpha[/tex] = level of significance = 100 - 90% = 10% = 0..10
[tex]\hat{p}[/tex] = sample proportion = ratio of 86.6% of n to n (at the least)

Part (a):

The number of subjects used at least one prescription medication is:

[tex]\dfrac{3149}{100} \times 86.6 \approx 2727[/tex]

Thus, the sample proportion we get is:

[tex]\hat{p} = \dfrac{2727}{3149} \approx 0.8659[/tex]

For level of significance 0.10, we get: [tex]Z_{\alpha/2} = 1.645[/tex]

Thus, the confidence interval needed is:

[tex]CI = \hat{p} \pm Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\CI \approx 0.8659\pm 1.645 \times \sqrt{\dfrac{0.8659(1-0.8659)}{3149}}\\\\\\CI \approx 0.8659 \pm 0.0099[/tex]

Thus, CI is [0.8659 - 0.0099, 0.8659 + 0.0099] = [0.8560, 0.8758]

Thus, the number of subjects in the study who used at least one prescription medication is 2727 approx. The needed 90% confidence interval is: [0.8560, 0.8758] or in percentage as: 85.6% to 87.58%

Learn more about population proportion here:

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